# Thread: laplace equation

1. ## laplace equation

I have to solve this:

$\Delta u(x, y)=0$
$u(0, y)=0$
$u(x,0)=1$

I've tried separation of variables, $u(x, y)=X(x)*Y(y)$ and got $X=C*sin \lambda x$
What next?
Is that even the right way to go?
Do I have to homogenize the conditions first?

Thank you.

2. Originally Posted by georgel
I have to solve this:

$\Delta u(x, y)=0$
$u(0, y)=0$
$u(x,0)=1$

I've tried separation of variables, $u(x, y)=X(x)*Y(y)$ and got $X=C*sin \lambda x$
What next?
Is that even the right way to go?
Do I have to homogenize the conditions first?

Thank you.
Using seperation of variables you should get $\frac{1}{Y} \frac{d^2Y}{dy^2} = - \frac{1}{X} \frac{d^2X}{dx^2}$.

Therefore:

$\frac{1}{Y} \frac{d^2Y}{dy^2} = k \Rightarrow \frac{d^2Y}{dy^2} - kY = 0$.

$\frac{1}{X} \frac{d^2X}{dx^2} = -k \Rightarrow \frac{d^2X}{dx^2} + kX = 0$.

Case 1: k = 0

Case 2: k < 0

Case 3: k > 0

Two of these cases lead to trivial solutions. It's the non-trivial one you need to work through.

3. Originally Posted by mr fantastic
I just realised .... you meant $\nabla u = 0$ .....

Using seperation of variables you should get $\frac{1}{Y} \frac{dY}{dy} = - \frac{1}{X} \frac{dX}{dx}$.

Therefore:

$\frac{1}{Y} \frac{dY}{dy} = k \Rightarrow \frac{dY}{dy} - kY = 0$.

$\frac{1}{X} \frac{dX}{dx} = -k \Rightarrow \frac{dX}{dx} + kX = 0$.

Case 1: k = 0

Case 2: k < 0

Case 3: k > 0

Two of these cases lead to trivial solutions. It's the non-trivial one you need to work through.
I think you meant to write,
$Y '' - kY = 0$ and $X''+ kX = 0$.

4. Originally Posted by ThePerfectHacker
I think you meant to write,
$Y '' - kY = 0$ and $X''+ kX = 0$.
After re-reading the original post just a bit more carefully, yes I did. Thanks for the catch.

Edit: But now after reading more of the thread I'm not sure that I did ......

5. First of all, thank you for your responses.

This is what I did:

$u(x, y)=X(x)Y(y)$
$X''Y + Y''X=0$
$\frac{X''}{X}=-\frac{Y''}{Y}=-\lambda^2$, $\lambda \geq 0$

According to my textbook, we use $\lambda$ _squared_ because it's easier to work with it later, and the minus is according to the theorem of non-negativity of eigenvalues (I'm not sure that's the correct english term.) so basically, I'm checking $\lambda =0$ and $\lambda > 0$.

$\lambda=0$ gives me:
$X''=0$, that is, $X=Ax+B$.
$X(0)=0$ gives me $B=0$.
So, I have $X=Ax$, which is not a trivial solution. I get the same when I do it like you said, $X''+kX=0$.
What did I do wrong?

$\lambda>0$:
Characteristic equation:
$x^2=-\lambda^2$
$x=+- \sqrt{-\lambda^2}=+-i\lambda$
$X=Csin(\lambda x)+Dcos(\lambda x)$
$X(0)=0$ gives me $D=0$. So, $X=Csin(\lambda x)$.

6. Originally Posted by georgel
First of all, thank you for your responses.

This is what I did:

$u(x, y)=X(x)Y(y)$
$X''Y + Y''X=0$
$\frac{X''}{X}=-\frac{Y''}{Y}=-\lambda^2$, $\lambda \geq 0$

According to my textbook, we use $\lambda$ _squared_ because it's easier to work with it later, and the minus is according to the theorem of non-negativity of eigenvalues (I'm not sure that's the correct english term.) so basically, I'm checking $\lambda =0$ and $\lambda > 0$.

$\lambda=0$ gives me:
$X''=0$, that is, $X=Ax+B$.
$X(0)=0$ gives me $B=0$.
So, I have $X=Ax$, which is not a trivial solution. I get the same when I do it like you said, $X''+kX=0$.
What did I do wrong?

$\lambda>0$:
Characteristic equation:
$x^2=-\lambda^2$
$x=+- \sqrt{-\lambda^2}=+-i\lambda$
$X=Csin(\lambda x)+Dcos(\lambda x)$
$X(0)=0$ gives me $D=0$. So, $X=Csin(\lambda x)$.

OK, now I'm confused. I might have let TPH change my mind incorrectly.

By $\Delta U = 0$ do you mean:

1. $\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial y^2} = 0$, or

2. $\nabla U = 0 \Rightarrow \frac{\partial U}{\partial x} + \frac{\partial U}{\partial y} = 0$.

I thought you meant 2. since there are only two boundary conditions given. But I changed my mind after reading TPH's post (after all, $\Delta$ is a commonly used symbol for the Laplacian operator).

So which is it? If it's 1. then there are no trivial solutions because there aren't enough boundary conditions to get rid of all the arbitrary constants.

If it's 2. I'll reply later since i have things to do at the moment.

7. I meant: $\Delta u= u_{xx} + u_{yy}=0$.

We've had a several exercises like this one in class, but there's always been more boundary conditions, and that's what's confusing me.

I've found a similar exercise in another textbook, but there it says to solve it on $R^+ \times R^+$. But even with that information, I still don't know what to do next..

8. Originally Posted by georgel
$\lambda=0$ gives me:
$X''=0$, that is, $X=Ax+B$.
$X(0)=0$ gives me $B=0$.
So, I have $X=Ax$, which is not a trivial solution. I get the same when I do it like you said, $X''+kX=0$.
What did I do wrong?
Here your equations lead to: $X'' = 0 \mbox{ and }Y'' = 0$.
Thus, $X = (Ax+B) \mbox{ and }Y = (Cy+D)$.
This means, $u(x,y) = XY = (Ax+B)(Cy+D)$.
Note $u(0,y)=0$ which means $(B)(Cy+D) = 0$.
To have non-trivial solutions we require $B=0$.
And $u(x,0)=1$ which means $(Ax)(D) = 1$.
But this is impossible because LHS is non constant.

Originally Posted by mr fantastic
OK, now I'm confused. I might have let TPH change my mind incorrectly.

By $\Delta U = 0$ do you mean:

1. $\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial y^2} = 0$, or

2. $\nabla U = 0 \Rightarrow \frac{\partial U}{\partial x} + \frac{\partial U}{\partial y} = 0$.
It means #1, there is no need to solve #2 by seperation of variables.
Because #2 has a general solution $u=f(x-y)$ where $f$ is a differenciable function.

9. Originally Posted by ThePerfectHacker
Here your equations lead to: $X'' = 0 \mbox{ and }Y'' = 0$.
Thus, $X = (Ax+B) \mbox{ and }Y = (Cy+D)$.
This means, $u(x,y) = XY = (Ax+B)(Cy+D)$.
Note $u(0,y)=0$ which means $(B)(Cy+D) = 0$.
To have non-trivial solutions we require $B=0$.
And $u(x,0)=1$ which means $(Ax)(D) = 1$.
But this is impossible because LHS is non constant.
It was incredibly stupid of me not to take into account the second equation.
Thank you very much!

So, I get that $X=sin \lambda x$, and that $Y=A e^ {\lambda y} + B e^{-\lambda y}$, where $A+B=1$.

I'll use it in the $\Delta u=0$ and hope something will work out. :-)

Thank you, both of you!